What is the solution of the equation - ln-dy-dx-x-0- where c is an arbitrary constant- y-e-x-c- y-e-x-c- y-e-x-c- y-e-x-c-

The correct option is C y+e^ x=c Given Differential equation is ln( dy dx #160; ) #160; +x=0 ⟹ dy dx = e ^ x Separating the variables ...

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Standard XII

Mathematics

Solving Linear Differential Equations of First Order

What is the s...

Question

What is the solution of the equation $l n (\frac{d y}{d x}) + x = 0$ ? where c is an arbitrary constant.

y + e^{x} = c

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y - e^{- x} = c

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y + e^{- x} = c

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y - e^{x} = c

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Solution

The correct option is C $y + e^{- x} = c$
Given Differential equation is $ln (\frac{d y}{d x}) + x = 0 ⟹ \frac{d y}{d x} = e^{- x}$
Separating the variables and integrating them gives,
$\int e^{- x} d x = \int d y$
$⟹ - e^{- x} = y + c$ Where $c$ is the arbitrary constant
$⟹ y + e^{- x} = c$ is the general solution and $c$ is the arbitrary constant.

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What is the solution of the equation ln(dydx)+x=0? where c is an arbitrary constant.

The correct option is C y+e−x=cGiven Differential equation is ln(dydx)+x=0⟹dydx=e−xSeparating the variables and integrating them gives,∫e−xdx=∫dy⟹−e−x=y+c Where c is the arbitrary constant⟹y+e−x=c is the general solution and c is the arbitrary constant.

What is the solution of the equation $l n (\frac{d y}{d x}) + x = 0$ ? where c is an arbitrary constant.